Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Ex 6.3, 8 Find a point on the curve π¦=(π₯β2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).Given Curves is π¦=(π₯β2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve π¦=(π₯β2)^2 Given that tangent is Parallel to the chord i.e. CD β₯ AB β΄ Slope of CD = Slope of AB If two lines are parallel, then their slopes are equal Slope of tangent CD Slope of tangent CD = ππ¦/ππ₯ =(π(π₯ β 2)^2)/ππ₯ = 2(π₯β2) (π (π₯ β 2))/ππ₯ = 2(π₯β2) (1β0) = 2(π₯β2) Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 β 0)/(4 β 2) =4/2 As slope of line joining point (π₯ , π¦) & (π₯2 , π¦2) ππ  (π¦2 β π¦1)/(π₯2 β π₯1) =2 Now, Slope of CD = Slope of AB 2(π₯β2)=2 π₯β2=2/2 π₯β2=1 π₯=3 Finding y when π₯=3 π¦=(π₯β2)^2 π¦=(3β2)^2 π¦=(1)^2 π¦=1 Hence, Point is (π , π) Thus, the tangent is parallel to the chord at (3 ,1)